Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)
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Transcript of Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)
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Sec on 3.7Indeterminate forms and lHôpital’s
Rule
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
March 30, 2011
Announcements
I Midterm has beenreturned. Please see FAQon Blackboard (under”Exams and Quizzes”)
I Quiz 3 this week inrecita on on Sec on 2.6,2.8, 3.1, 3.2
ObjectivesI Know when a limit is ofindeterminate form:
I indeterminate quo ents:0/0,∞/∞
I indeterminate products:0×∞
I indeterminatedifferences: ∞−∞
I indeterminate powers:00,∞0, and 1∞
I Resolve limits inindeterminate form
Recall
Recall the limit laws from Chapter 2.I Limit of a sum is the sum of the limits
I Limit of a difference is the difference of the limitsI Limit of a product is the product of the limitsI Limit of a quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.
Recall
Recall the limit laws from Chapter 2.I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limits
I Limit of a product is the product of the limitsI Limit of a quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.
Recall
Recall the limit laws from Chapter 2.I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limitsI Limit of a product is the product of the limits
I Limit of a quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.
Recall
Recall the limit laws from Chapter 2.I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limitsI Limit of a product is the product of the limitsI Limit of a quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.
More about dividing limits
I We know dividing by zero is bad.I Most of the me, if an expression’s numerator approaches afinite nonzero number and denominator approaches zero, thequo ent has an infinite. For example:
limx→0+
1x= +∞ lim
x→0−
cos xx3
= −∞
Why 1/0 ̸= ∞Consider the func on f(x) =
11x sin x
.
..x
.
y
Then limx→∞
f(x) is of the form 1/0, but the limit does not exist and isnot infinite.
Even less predictable: when numerator and denominator both go tozero.
Why 1/0 ̸= ∞Consider the func on f(x) =
11x sin x
.
..x
.
y
Then limx→∞
f(x) is of the form 1/0, but the limit does not exist and isnot infinite.Even less predictable: when numerator and denominator both go tozero.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
Language NoteIt depends on what the meaning of the word “is” is
I Be careful with the language here. Weare not saying that the limit in eachcase “is”
00, and therefore nonexistent
because this expression is undefined.
I The limit is of the form00, which means
we cannot evaluate it with our limitlaws.
Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
The Linear CaseQues on
If f and g are lines and f(a) = g(a) = 0, what is limx→a
f(x)g(x)
?
Solu onThe func ons f and g can be wri en in the form
f(x) = m1(x− a) g(x) = m2(x− a)
Sof(x)g(x)
=m1
m2
The Linear CaseQues on
If f and g are lines and f(a) = g(a) = 0, what is limx→a
f(x)g(x)
?
Solu onThe func ons f and g can be wri en in the form
f(x) = m1(x− a) g(x) = m2(x− a)
Sof(x)g(x)
=m1
m2
The Linear Case, Illustrated
.. x.
y
.
y = f(x)
.
y = g(x)
..a
..x
.f(x)
.g(x)
f(x)g(x)
=f(x)− f(a)g(x)− g(a)
=(f(x)− f(a))/(x− a)(g(x)− g(a))/(x− a)
=m1
m2
What then?
I But what if the func ons aren’t linear?
I Can we approximate a func on near a point with a linearfunc on?
I What would be the slope of that linear func on?
Thederiva ve!
What then?
I But what if the func ons aren’t linear?I Can we approximate a func on near a point with a linearfunc on?
I What would be the slope of that linear func on?
Thederiva ve!
What then?
I But what if the func ons aren’t linear?I Can we approximate a func on near a point with a linearfunc on?
I What would be the slope of that linear func on?
Thederiva ve!
What then?
I But what if the func ons aren’t linear?I Can we approximate a func on near a point with a linearfunc on?
I What would be the slope of that linear func on? Thederiva ve!
Theorem of the DayTheorem (L’Hopital’s Rule)
Suppose f and g are differen able func ons and g′(x) ̸= 0 near a(except possibly at a). Suppose that
limx→a
f(x) = 0 and limx→a
g(x) = 0
or limx→a
f(x) = ±∞ and limx→a
g(x) = ±∞
Thenlimx→a
f(x)g(x)
= limx→a
f′(x)g′(x)
,
if the limit on the right-hand side is finite,∞, or−∞.
Meet the Mathematician
I wanted to be a military man, butpoor eyesight forced him intomath
I did some math on his own(solved the “brachistocroneproblem”)
I paid a s pend to JohannBernoulli, who proved thistheorem and named it a er him!
Guillaume François Antoine,Marquis de L’Hôpital(French, 1661–1704)
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
..
sin x → 0
cos x1
= 0
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
..
sin x → 0
cos x1
= 0
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x..
sin x → 0
cos x1
= 0
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x..
sin x → 0
cos x1
= 0
Revisiting the previous examples
Example
limx→0
sin2 x
..
numerator → 0
sin x2
..
denominator → 0
H= lim
x→0
��2 sin x cos x
..
numerator → 0
(cos x2) (��2x
..
denominator → 0
)H= lim
x→0
cos2 x− sin2 x
..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x..
numerator → 0
sin x2
..
denominator → 0
H= lim
x→0
��2 sin x cos x
..
numerator → 0
(cos x2) (��2x
..
denominator → 0
)H= lim
x→0
cos2 x− sin2 x
..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x..
numerator → 0
sin x2..
denominator → 0
H= lim
x→0
��2 sin x cos x
..
numerator → 0
(cos x2) (��2x
..
denominator → 0
)H= lim
x→0
cos2 x− sin2 x
..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x..
numerator → 0
sin x2..
denominator → 0
H= lim
x→0
��2 sin x cos x
..
numerator → 0
(cos x2) (��2x
..
denominator → 0
)
H= lim
x→0
cos2 x− sin2 x
..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x
..
numerator → 0
sin x2
..
denominator → 0
H= lim
x→0
��2 sin x cos x..
numerator → 0
(cos x2) (��2x
..
denominator → 0
)
H= lim
x→0
cos2 x− sin2 x
..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x
..
numerator → 0
sin x2
..
denominator → 0
H= lim
x→0
��2 sin x cos x..
numerator → 0
(cos x2) (��2x..
denominator → 0
)
H= lim
x→0
cos2 x− sin2 x
..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x
..
numerator → 0
sin x2
..
denominator → 0
H= lim
x→0
��2 sin x cos x..
numerator → 0
(cos x2) (��2x..
denominator → 0
)H= lim
x→0
cos2 x− sin2 x
..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x
..
numerator → 0
sin x2
..
denominator → 0
H= lim
x→0
��2 sin x cos x
..
numerator → 0
(cos x2) (��2x
..
denominator → 0
)H= lim
x→0
cos2 x− sin2 x..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x
..
numerator → 0
sin x2
..
denominator → 0
H= lim
x→0
��2 sin x cos x
..
numerator → 0
(cos x2) (��2x
..
denominator → 0
)H= lim
x→0
cos2 x− sin2 x..
numerator → 1
cos x2 − 2x2 sin(x2)..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin2 x
..
numerator → 0
sin x2
..
denominator → 0
H= lim
x→0
��2 sin x cos x
..
numerator → 0
(cos x2) (��2x
..
denominator → 0
)H= lim
x→0
cos2 x− sin2 x
..
numerator → 1
cos x2 − 2x2 sin(x2)
..
denominator → 1
= 1
Revisiting the previous examples
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
Revisiting the previous examples
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
Another Example
Example
Findlimx→0
xcos x
Solu onThe limit of the denominator is 1, not 0, so L’Hôpital’s rule does notapply. The limit is 0.
Beware of Red Herrings
Example
Findlimx→0
xcos x
Solu onThe limit of the denominator is 1, not 0, so L’Hôpital’s rule does notapply. The limit is 0.
Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
Limits of Rational Functionsrevisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
Limits of Rational Functionsrevisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
Limits of Rational Functionsrevisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
Limits of Rational Functionsrevisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
Limits of Rational Functionsrevisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
Limits of Rational Functionsrevisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
Limits of Rational Functionsrevisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
Limits of Rational Functionsrevisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
Limits of Rational Functionsrevisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
Limits of Rational Functionsrevisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
Limits of Rational Functionsrevisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
Limits of Rational Functionsrevisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
Limits of Rational Functionsrevisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
Limits of Rational FunctionsFactLet f(x) and g(x) be polynomials of degree p and q.
I If p > q, then limx→∞
f(x)g(x)
= ∞
I If p < q, then limx→∞
f(x)g(x)
= 0
I If p = q, then limx→∞
f(x)g(x)
is the ra o of the leading coefficients of
f and g.
Exponential vs. geometric growth
Example
Find limx→∞
ex
x2, if it exists.
Solu onWe have
limx→∞
ex
x2H= lim
x→∞
ex
2xH= lim
x→∞
ex
2= ∞.
Exponential vs. geometric growth
Example
Find limx→∞
ex
x2, if it exists.
Solu onWe have
limx→∞
ex
x2H= lim
x→∞
ex
2xH= lim
x→∞
ex
2= ∞.
Exponential vs. geometric growthExample
What about limx→∞
ex
x3?
AnswerS ll∞. (Why?)
Solu on
limx→∞
ex
x3H= lim
x→∞
ex
3x2H= lim
x→∞
ex
6xH= lim
x→∞
ex
6= ∞.
Exponential vs. geometric growthExample
What about limx→∞
ex
x3?
AnswerS ll∞. (Why?)
Solu on
limx→∞
ex
x3H= lim
x→∞
ex
3x2H= lim
x→∞
ex
6xH= lim
x→∞
ex
6= ∞.
Exponential vs. geometric growthExample
What about limx→∞
ex
x3?
AnswerS ll∞. (Why?)
Solu on
limx→∞
ex
x3H= lim
x→∞
ex
3x2H= lim
x→∞
ex
6xH= lim
x→∞
ex
6= ∞.
Exponential vs. fractional powersExample
Find limx→∞
ex√x, if it exists.
Exponential vs. fractional powersExample
Find limx→∞
ex√x, if it exists.
Solu on (without L’Hôpital)
We have for all x > 1, x1/2 < x1, so
ex
x1/2>
ex
x
The right hand side tends to∞, so the le -hand side must, too.
Exponential vs. fractional powersExample
Find limx→∞
ex√x, if it exists.
Solu on (with L’Hôpital)
limx→∞
ex√x= lim
x→∞
ex12x−1/2 = lim
x→∞2√xex = ∞
Exponential vs. any powerTheorem
Let r be any posi ve number. Then limx→∞
ex
xr= ∞.
Proof.
Exponential vs. any powerTheorem
Let r be any posi ve number. Then limx→∞
ex
xr= ∞.
Proof.If r is a posi ve integer, then apply L’Hôpital’s rule r mes to the frac-on. You get
limx→∞
ex
xrH= . . .
H= lim
x→∞
ex
r!= ∞.
Exponential vs. any powerTheorem
Let r be any posi ve number. Then limx→∞
ex
xr= ∞.
Proof.If r is not an integer, letm be the smallest integer greater than r. Thenif x > 1, xr < xm, so
ex
xr>
ex
xm. The right-hand side tends to∞ by the
previous step.
Any exponential vs. any powerTheorem
Let a > 1 and r > 0. Then limx→∞
ax
xr= ∞.
Proof.If r is a posi ve integer, we have
limx→∞
ax
xrH= . . .
H= lim
x→∞
(ln a)rax
r!= ∞.
If r isn’t an integer, we can compare it as before.
So even limx→∞
(1.00000001)x
x100000000= ∞!
Any exponential vs. any powerTheorem
Let a > 1 and r > 0. Then limx→∞
ax
xr= ∞.
Proof.If r is a posi ve integer, we have
limx→∞
ax
xrH= . . .
H= lim
x→∞
(ln a)rax
r!= ∞.
If r isn’t an integer, we can compare it as before.
So even limx→∞
(1.00000001)x
x100000000= ∞!
Any exponential vs. any powerTheorem
Let a > 1 and r > 0. Then limx→∞
ax
xr= ∞.
Proof.If r is a posi ve integer, we have
limx→∞
ax
xrH= . . .
H= lim
x→∞
(ln a)rax
r!= ∞.
If r isn’t an integer, we can compare it as before.
So even limx→∞
(1.00000001)x
x100000000= ∞!
Logarithmic versus power growthTheorem
Let r be any posi ve number. Then limx→∞
ln xxr
= 0.
Proof.One applica on of L’Hôpital’s Rule here suffices:
limx→∞
ln xxr
H= lim
x→∞
1/xrxr−1 = lim
x→∞
1rxr
= 0.
Logarithmic versus power growthTheorem
Let r be any posi ve number. Then limx→∞
ln xxr
= 0.
Proof.One applica on of L’Hôpital’s Rule here suffices:
limx→∞
ln xxr
H= lim
x→∞
1/xrxr−1 = lim
x→∞
1rxr
= 0.
Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
Indeterminate productsExample
Find limx→0+
√x ln x
This limit is of the form 0 · (−∞).
Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:
limx→0+
√x ln x
= limx→0+
ln x1/√x
H= lim
x→0+
x−1
−12x−3/2 = lim
x→0+−2
√x = 0
Indeterminate productsExample
Find limx→0+
√x ln x
This limit is of the form 0 · (−∞).
Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:
limx→0+
√x ln x
= limx→0+
ln x1/√x
H= lim
x→0+
x−1
−12x−3/2 = lim
x→0+−2
√x = 0
Indeterminate productsExample
Find limx→0+
√x ln x
This limit is of the form 0 · (−∞).
Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+
ln x1/√x
H= lim
x→0+
x−1
−12x−3/2 = lim
x→0+−2
√x = 0
Indeterminate productsExample
Find limx→0+
√x ln x
This limit is of the form 0 · (−∞).
Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+
ln x1/√x
H= lim
x→0+
x−1
−12x−3/2
= limx→0+
−2√x = 0
Indeterminate productsExample
Find limx→0+
√x ln x
This limit is of the form 0 · (−∞).
Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+
ln x1/√x
H= lim
x→0+
x−1
−12x−3/2 = lim
x→0+−2
√x
= 0
Indeterminate productsExample
Find limx→0+
√x ln x
This limit is of the form 0 · (−∞).
Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+
ln x1/√x
H= lim
x→0+
x−1
−12x−3/2 = lim
x→0+−2
√x = 0
Indeterminate differences
Example
limx→0+
(1x− cot 2x
)This limit is of the form∞−∞.
Indeterminate DifferencesSolu onAgain, rig it to make an indeterminate quo ent.
limx→0+
(1x− cot 2x
)= lim
x→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+
cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.
Indeterminate DifferencesSolu onAgain, rig it to make an indeterminate quo ent.
limx→0+
(1x− cot 2x
)= lim
x→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+
cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.
Indeterminate DifferencesSolu onAgain, rig it to make an indeterminate quo ent.
limx→0+
(1x− cot 2x
)= lim
x→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+
cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.
Indeterminate DifferencesSolu onAgain, rig it to make an indeterminate quo ent.
limx→0+
(1x− cot 2x
)= lim
x→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+
cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.
Checking your workThis all goes in the thought cloud
limx→0
tan 2x2x
= 1, so for small x, tan 2x ≈ 2x. So cot 2x ≈ 12x
and
1x− cot 2x ≈ 1
x− 1
2x=
12x
→ ∞
as x → 0+.
Indeterminate powersExample
Find limx→0+
(1− 2x)1/x
Solu on
Indeterminate powersExample
Find limx→0+
(1− 2x)1/x
Solu onTake the logarithm:
ln(
limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+
ln(1− 2x)x
Indeterminate powersExample
Find limx→0+
(1− 2x)1/x
Solu on
This limit is of the form00, so we can use L’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x
1= −2
Indeterminate powersExample
Find limx→0+
(1− 2x)1/x
Solu on
This limit is of the form00, so we can use L’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x
1= −2
This is not the answer, it’s the log of the answer! So the answer wewant is e−2.
Another indeterminate power limitExample
Find limx→0+
(3x)4x
Solu on
ln limx→0+
(3x)4x = limx→0+
ln(3x)4x = limx→0+
4x ln(3x) = limx→0+
ln(3x)1/4x
H= lim
x→0+
3/3x−1/4x2
= limx→0+
(−4x) = 0
So the answer is e0 = 1.
Another indeterminate power limitExample
Find limx→0+
(3x)4x
Solu on
ln limx→0+
(3x)4x = limx→0+
ln(3x)4x = limx→0+
4x ln(3x) = limx→0+
ln(3x)1/4x
H= lim
x→0+
3/3x−1/4x2
= limx→0+
(−4x) = 0
So the answer is e0 = 1.
SummaryForm Method
00 L’Hôpital’s rule directly∞∞ L’Hôpital’s rule directly
0 · ∞ jiggle to make 00 or
∞∞ .
∞−∞ combine to make an indeterminate product or quo ent
00 take ln to make an indeterminate product
∞0 di o
1∞ di o
Final Thoughts
I L’Hôpital’s Rule only works on indeterminate quo entsI Luckily, most indeterminate limits can be transformed intoindeterminate quo ents
I L’Hôpital’s Rule gives wrong answers for non-indeterminatelimits!